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cho 3 số a, b, c hác 0 thỏa mãn ab/ (a+b) = bc/ (b+c) = ca/ (c+a)
Tính M = ab + bc + ca/ a2 + b2 + c2
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
Tính M = ab + bc + ca/ a2 + b2 + c2
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}=\frac{1}{c}\\\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\Rightarrow\frac{1}{b}=\frac{1}{a}\\\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{a}=\frac{1}{c}=\frac{1}{a}\end{cases}}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
\(\Rightarrow M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{1.1+1.1+1.1}{1^2+1^2+1^2}=\frac{3}{3}=1\)
Ta có \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
Mà \(a,b,c \ne0\) => \(ab,bc,ca \ne0\)
=> \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
=> \(\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
=> \(\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
=> \(a=b=c\)
Thay vào M ta có : \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a.a+a.a+a.a}{a^2+a^2+a^2}=\frac{3a^2}{3a^2}=1\)
Vậy \(M=1\)
Câu hỏi của Đậu Đình Kiên - Toán lớp 7 - Học toán với OnlineMath
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ac}{a+c}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{a+c}{ac}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{a}{ac}+\frac{c}{ac}\)
\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{c}+\frac{1}{a}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}\\\frac{1}{c}+\frac{1}{b}=\frac{1}{c}+\frac{1}{a}\\\frac{1}{c}+\frac{1}{a}=\frac{1}{b}+\frac{1}{a}\end{cases}}\) \(\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{a}\\\frac{1}{c}=\frac{1}{b}\end{cases}}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
\(\Rightarrow a=b=c\)
Khi đó : \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{1.1+1.1+1.1}{1^2+1^2+1^2}=\frac{3}{3}=1\)
Vậy \(M=1\)
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{a+c}\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ac}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\Rightarrow M=1\)
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\Rightarrow\frac{1}{a}=\frac{1}{c}\left(1\right)\)
\(\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Rightarrow\frac{1}{b}=\frac{1}{a}\left(2\right)\)
Từ (1) và (2) => \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Vậy \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
\(\frac{a.b}{a+b}=\frac{b.c}{b+c}=\frac{c.a}{c+a}\)
\(\Rightarrow\frac{a+b}{a.b}=\frac{b+c}{b.c}=\frac{c+a}{c.a}\) (vì a;b;c khác 0)
\(=\frac{a}{a.b}+\frac{b}{a.b}=\frac{b}{b.c}+\frac{c}{b.c}=\frac{c}{c.a}+\frac{a}{c.a}\)
\(=\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
=> a = b = c
\(P=\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\frac{a.a^2+a.a^2+a.a^2}{a^3+a^3+a^3}=\frac{a^3+a^3+a^3}{a^3+a^3+a^3}=1\)
Câu hỏi của Đậu Đình Kiên - Toán lớp 7 - Học toán với OnlineMath